JEE Main 2019 — Modern Physics

Light is incident normally on a completely absorbing surface with an energy flux of $25 W c{m}^{-2}$ . If the surface has an area of $25 c{m}^{2}$ , the momentum transferred to the surface in $\text{40} \text{min}$ time duration will be:

$Ge$ and $Si$ diodes start conducting at $0.3 V$ and $0.7 V$ respectively. In the following figure if $Ge$ diode connection are reversed, the value of ${V}_{0}$ changes by: (assume that the Ge diode has large breakdown voltage) 

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths $\frac{{\lambda }_{1}}{{\lambda }_{2}}$ of the photons emitted in this process is:

When a certain photosensitive surface is illuminated with a monochromatic light of frequency $\nu ,$ the stopping potential of the photo current is $-\frac{{V}_{0}}{2}.$ When the surface is illuminated by monochromatic light of frequency $\frac{\nu }{2},$ the stopping potential is $-{V}_{0}.$ The threshold frequency for photoelectric emission is

Two particles move at right angle to each other. Their de Broglie wavelengths are ${\lambda }_{1}$ and ${\lambda }_{2}$ respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle, is given by:

A nucleus $A,$ with a finite de-broglie wavelength ${\lambda }_{A},$ undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A,$ while $C$ flies in the opposite direction with a velocity equal to half of that of $B.$ The de-Broglie wavelengths ${\lambda }_{B}$ and ${\lambda }_{C}$ of B and C are respectively:

To get output '1' at $R,$ for the given logic gate circuit the input values must be: 

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is $6 V$ and the load resistance is, ${R}_{L}=4 k\Omega$ . The series resistance of the circuit is ${R}_{i}=1 k\Omega$ . If the battery voltage ${V}_{B}$ varies from $8 V$ to $16 V$ , what are the minimum and maximum values of the current through Zener diode? 

A particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 V.$ Another particle $B$ of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500 V.$ The ratio of de-Broglie wavelengths $\frac{{\lambda }_{A}}{{\lambda }_{B}}$ is close to:

Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on ${He}^{+}$ ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation is:

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $L$. If an electron jumps from $\mathrm{N}$ -shell to the $\mathrm{L}$ -shell, the wavelength of emitted radiation will be:

The reverse break down voltage of a Zener diode is $5.6 V$ in the given circuit.  The current ${I}_{z}$ through the Zener is:

The logic gate equivalent to the given logic circuit is: 

A $2 mW$ laser operates at a wavelength of $500 nm$ . The number of photons that will be emitted per second is: [Given Planck's constant $h=6.6\times {10}^{-34} J s,$ speed of light $c=3.0\times {10}^{8} m/s$ ]

If the deBroglie wavelength of an electron is equal to $10^{-}$ times the wavelength of a photon of frequency $6 \times 10^{14}$ $\mathrm{Hz}$, then the speed of electron is equal to : (Speed of light $\left.=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ Mass of electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)$

The magnetic field associated with a light wave is given, at the origin, by $B={B}_{0} [\mathrm{sin}(3.14\times {10}^{7})ct+\mathrm{sin}(6.28\times {10}^{7})ct].$ If this light falls on a silver plate having a work function of $4.7\mathrm{eV},$ what will be the maximum kinetic energy of the photoelectrons? $(c=3\times {10}^{8} m {s}^{-1},h=6.6\times {10}^{-34} J s)$

In a Frank - Hertz experiment, an electron of energy $5.6 eV$ passes through mercury vapour and emerges with an energy $0.7 eV.$ The minimum wavelength of photons emitted by mercury atoms is close to:

In$L{i}^{++}$,electron in first Bohr orbit is excited to a level by a radiation of wavelength$\lambda$.When the ion gets de excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$ ? (Given: $h=6.63\times {10}^{-34} J s ;c=3\times {10}^{8} m {s}^{-1}$ )

Figure shows a $DC$ voltage regulator circuit, with a Zener diode of breakdown voltage $=6 V$ . If the unregulated input voltage varies between $10 V$ to $16 V,$ then what is the maximum Zener current? 

The stopping potential ${V}_{o}$ (in volt) as a function of frequency $(v)$ for a sodium emitter, is shown in the figure. The work function of sodium, form the data plotted in the figure, will be: (Given: Planck’s constant $(h)=6.63\times {10}^{-34} J s$ , electron charge $( \text{e} )$ $=1.6\times {10}^{-19} C)$ 

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 \mathrm{nm}$ to $400 \mathrm{nm}$. The decrease in the stopping potential is close to : $\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\right)$

For the circuit shown below, the current through the Zener diode is 

Consider the nuclear fission, ${\mathrm{Ne}}^{20}\rightarrow 2{\mathrm{He}}^{4}+{C}^{12}$. Given that the binding energy/nucleon of ${\mathrm{Ne}}^{20},{\mathrm{He}}^{4}$and${C}^{12}$ are $8.03 \mathrm{MeV},7.86\mathrm{MeV}$, respectively. Identify the correct statement:

In a photoelectric effect experiment, the threshold wavelength of light is$380 nm$ . If the wavelength of incident light is$260 nm$ , the maximum kinetic energy of emitted electrons will be Given $E$ (in $eV$ ) $=\frac{1237}{\lambda (in nm)}$

The ratio of mass densities of nuclei of ${ }^{40}Ca$ and${ }^{16}O$ is close to:

An excited ${He}^{+}$ ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$ in making a transition to the ground state. The quantum number $n$, corresponding to its initial excited state is (for a photon of wavelength $\lambda$, energy $E=\frac{1240\mathrm{eV}}{\lambda (\mathrm{in} \mathrm{nm})}$)

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 Ã…$. The radius of the atom in the excited state, in terms of Bohr radius $a_{0}$, will be:

A particle of mass $m$ moves in a circular orbit in a central potential field $U(r)=\frac{1}{2}k{r}^{2}.$ If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as:

A particle $'P'$ is formed due to a completely inelastic collision of particles $'x'$ and $'y'$ having de-Broglie wavelengths $'{\lambda }_{x}'$ and $'{\lambda }_{y}'$ respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of $'P'$ is:

The surface of certain metal is first illuminated with light of wavelength ${\lambda }_{1}=350 nm$ and then, by a light of wavelength ${\lambda }_{2}=540 nm.$ It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of $2$. The work function of the metal (in $eV$) is close to (Energy of photon $=\frac{1240}{\lambda (in nm)} eV$ )

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times {10}^{-12} m,$ the minimum electron energy required is close to:

Taking the wavelength of first Balmer line in hydrogen spectrum $(n=3$ to $n=2)$ as $660 nm$ , the wavelength of the ${2}^{nd}$ Balmer line $(n = 4$ to $n = 2)$ will be :

A metal plate of area $1\times {10}^{-4}{m}^{2}$ is illuminated by a radiation of intensity $16\frac{milliW}{{m}^{2}}$ . The work function of the metal is $5eV.$ The energy of the incident photons is $10eV$ and only $10%$ of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: $[1 eV=1.6\times {10}^{-19}J]$

The electric field of light wave is given as $\vec{E }={10}^{–3}\mathrm{cos}⁡(\frac{2\pi x}{5\times {10}^{–7}}-2\pi \times 6\times {10}^{14} t)\overset{^}{x}\frac{N}{C}$ . This light falls on a metal plate of work function $2 eV$ . The stopping potential of the photo-electrons is: Given, $E$ (in $eV$ ) $=\frac{12375}{\lambda (in Å)}$

The output of the given logic circuit is: 

The circuit shown below contains two ideal diodes, each with a forward resistance of $50 \Omega$. If the battery voltage is $6 \mathrm{~V},$ the current through the $100 \Omega$ resistance (in Amperes) is: 

In the given circuit the current through Zener Diode is close to: 

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \text{Å}$ ). The de-Broglie wavelength of this electron is:

A $H{e}^{+}$ ion is in its first excited state. Its ionization energy is:

The truth table for the circuit given in the figure is: